Let $n \in \mathbb N$, let $S_n$ be the symmetric group of bijections $\{1,\dots,n\}\to\{1,\dots,n\}$, and let $\operatorname{SYT}_n$ denote the set of standard Young tableaux with entries $\{1,\dots,n\}$. Let $\Lambda$ denote the Young lattice of partitions ordered by adding one box. For a saturated Young-lattice chain $\varnothing=\lambda_0\subset\lambda_1\subset\cdots\subset\lambda_n$ with $|\lambda_k|=k$, let $\operatorname{Tab}(\lambda_0,\dots,\lambda_n)\in\operatorname{SYT}_n$ be the tableau whose entry $k$ lies in the unique box of $\lambda_k\setminus\lambda_{k-1}$. Let $\operatorname{evac}_n:\operatorname{SYT}_n\to\operatorname{SYT}_n$ denote Schützenberger evacuation, defined by the usual jeu de taquin evacuation algorithm and equivalently by Fomin's reversible triangular evacuation growth diagram on the Young lattice: if $T=\operatorname{Tab}(\lambda_0,\dots,\lambda_n)$, then $\operatorname{evac}_n(T)$ is the tableau encoded by the opposite boundary chain produced from $(\lambda_0,\dots,\lambda_n)$ by the reversible Fomin local rule. Assume this triangular construction is unique from either boundary chain and that using the opposite boundary as input reconstructs the same triangular labelled diagram in reverse. Use the standard row-insertion Fomin local rule. For every permutation mark set $M\subset\{1,\dots,n\}\times\{1,\dots,n\}$ with one mark in each row and column, let $\Gamma_M:\{0,\dots,n\}\times\{0,\dots,n\}\to\Lambda$ be the unique empty-south-and-west-boundary Fomin growth diagram with mark set $M$. Its north boundary chain encodes the recording tableau and its east boundary chain encodes the insertion tableau of row-insertion RSK. Assume Fomin's square reversal theorem in this convention: rotating a finite empty-boundary permutation growth diagram by $180^\circ$, replacing the mark set by the rotated mark set, and reading the ordinary empty-boundary growth diagram for that rotated mark set sends the original northeast boundary tableaux to their Schützenberger evacuations. Then, for every $T \in \operatorname{SYT}_n$, $\operatorname{evac}_n(\operatorname{evac}_n(T))=T$. For $\pi \in S_n$, write $\pi=\pi_1\cdots\pi_n$ in one-line notation, where $\pi_i=\pi(i)$. Define the reverse-complement permutation $\pi^\vee \in S_n$ by $\pi^\vee(i)=n+1-\pi(n+1-i)$ for every $1 \le i \le n$. Let $\operatorname{RSK}_n:S_n\to\{(P,Q):P,Q\in\operatorname{SYT}_n\text{ have the same shape}\}$ be the row-insertion Robinson-Schensted correspondence, and write $\operatorname{RSK}_n(\pi)=(P(\pi),Q(\pi))$. Then $P(\pi^\vee)=\operatorname{evac}_n(P(\pi))$ and $Q(\pi^\vee)=\operatorname{evac}_n(Q(\pi))$.